Bond order of 1.5 is shown by

$\begin{aligned} & \text{(a) } \text {MO configuration of } \mathrm{O}_2^{+}(8+8-1=15) \\ & =\sigma 1 s^2, \sigma^* 1 s^2, \sigma 2 s^2, \stackrel{*}{\sigma} 2 s^2, \sigma 2 p_z^2, \\ & \pi 2 p_x^2 \approx \pi 2 p_y^2, \pi 2 p_x^1 \approx \pi 2 p_y^0 \\ & \mathrm{BO}=\frac{N_b-N_a}{2} \end{aligned}$ (where, $N_b=$ number of electrons in bonding molecular orbital $N_a=$ number of electrons in anti-bonding molecular orbital $\therefore \mathrm{BO}=\frac{10-5}{2}=2.5$ Similarly, (b) $\mathrm{O}_2^{-}(8+8+1=17)$ so $\mathrm{BO}=\frac{N_b-N_a}{2}=\frac{10-7}{2}=1.5$ (c) $\mathrm{O}_2^{2-}(8+8+2=18)$ $\mathrm{BO}=\frac{N_b-N_a}{2}=\frac{10-8}{2}=1$ (d) $\mathrm{O}_2(8+8=16)$ $\mathrm{BO}=\frac{10-6}{2}=2$ Thus, $\mathrm{O}_2^{-}$shows the bond order 1.5.